Linear Proof, don't know how to start it

[Dustinsfl]

Let A be nxn and let B=I-2A+A^2.
Show that if \mathbf{x} is an eigenvector of A belonging to an eigenvalue \lambda of A, then \mathbf{x} is also an eigenvector of B belonging to an eigenvalue \mu of B. How are \lambda and \mu related?

 

[nicksauce]

Well... what happens if you act 1 - 2A + A^2 on x?

 

[Dustinsfl]

Well... what happens if you act 1 - 2A + A^2 on x?

I don't know what you mean by that.

 

[nicksauce]

You want to show that x is an eigenvector B, meaning that for some constant u, Bx = ux.
Thus you need to calculate Bx = (1 -2A + A^2)x = x - 2Ax + A^2x. Can you simplify that further now?

 

[Dustinsfl]

B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mu\mathbf{x}

?

 

[Dick]

Evaluate A(lambda*x), please?

 

[Dustinsfl]

Evaluate A(lambda*x), please?

B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mathbf{x}-2\lambda\mathbf{x}+(\lambda)^2\mathbf{x}=\mu

 

[Dustinsfl]

Then do the quadratic equation or factor which ever is appropriate?

 

[Dick]

Then do the quadratic equation or factor which ever is appropriate?

You don't have to do much of anything. Just factor x out and say how Bx is related to mu*x.

 

[Dustinsfl]

Should I just solve for mu then by multiplying by 1/x?

 

[Dick]

Should I just solve for mu then by multiplying by 1/x?

No, there's no such thing as 1/x if x is a vector. That's exactly the wrong answer. Just look at x-2*lambda*x+lambda^2*x=mu*x=Bx and tell me what mu is.

 

[Dustinsfl]

So \mu=I-2\lambda+A\lambda.

 

[Dick]

So \mu=I-2\lambda+A\lambda.

mu is a scalar. It's a number. It's not a matrix. It's an EIGENVALUE. Try that again. You were almost there. Now you are going backwards.

 

[Dustinsfl]

mu is a scalar. It's a number. It's not a matrix. It's an EIGENVALUE. Try that again. You were almost there. Now you are going backwards.

\mu=1-2\lambda+A\lambda

But there is still A

 

[Dick]

The A is there because once upon a time I asked you to evaluate A(lambda*x). And you did it right. And then you forgot. Why did you do that? That's why I feel things are going backwards.

 

[Dustinsfl]

B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mathbf{x}-2\lambda\mathbf{x}+(\lambda)^2=\mu\mathbf{x}

I did. I just decide not use it apparently.

 

[Dick]

I did. I just decide not use it apparently.

Ok, so 1-2*lambda+lambda^2=mu. Right? I so very hope you agree with this.

 

[Dustinsfl]

I concur.